1) Suppose the counting numbers from 1 through 100 are written on paper. What is the total numbers of 3s and 8s that will appear on the paper?
2) In the "magic-square" below, five more numbers can be placed in the boxes so that the sum of the three numbers in each row, in each column, and in each diagonal is always the same. What value should X have? 3) The U-shaped figure at the right contains 11 squares of the same size. The area of the U-shaped figure is 176 square inches. How many inches are in the perimeter of the U-shaped figure? Solutions 1) (a) 3 occurs in the units place once in each group of ten consecutive numbers. Therefore 3 will appear 10 times in the units place. (b) 3 occurs in the tens place ten times in 100 consecutive numbers: 30, 31, 32, 33, . . . , 38, 39. (c) Statements (a) and (b) are also true for occurrences of 8. Therefore there will be a total of 40 occurrences of 3s and 8s in the numbers from 1 through 100. 2) Method 1 From the first column, we see that the sum for each row, column, and diagonal should be 90. Then the missing number in the 3rd row is 40, and the missing number in the lead diagonal is 30. The sum for the second column is 40 + 30 + X or 70 + X. Then X must be 20. Method 2 Since the numbers in the two diagonals have equal sums and the same middle number, then the two numbers in the corners of each diagonal have equal sums; 25 + 35 = 15 + ? It follows that ?, the number in the lower right corner, is 45. Since the sums of the numbers in any row, column, or diagonal is 90, the bottom row sum 25 + X + 45 = 90. Thus X = 20. 3) The area of each square is 176/11 or 16 inches. Then the length of each side of a square is 4 inches. The perimeter of the U-shaped figure is equivalent to the total length of 24 sides or 96 inches.
0 Comments
Leave a Reply. |
AuthorMrs. Lovett Archives
June 2019
Categories
All
|